Question

The air in a room with volume 180 m3 contains 0.4% carbon dioxide initially. Fresher air with only 0.05% carbon dioxide flows into the room at a rate of 2 m3/min and the mixed air flows out at the same rate. Find the percentage p of carbon dioxide in the room as a function of time t (in minutes).

p(t) = ___________

What happens with the percentage of carbon dioxide in the room in the long run?

lim t → [infinity] p(t) =________ %

Answer 1

Answer:

(A) The percentage p(t) of carbon dioxide in the room as a function of time t (in minutes) is $p(t)=0.55\%\left(0.63e^{-\frac{t}{90}}+0.09\right)$.

(B) The percentage of carbon dioxide will approach 0.0495%.

Step-by-step explanation:

Mixing problems are an application of separable differential equations. They’re word problems that require us to create a separable differential equation based on the concentration of a substance in a vessel.

Let y(t) be the total volume of carbon dioxide in the room.

The main equation that we’ll be using to model this situation is

Rate of change y(t) = Rate at which y(t) enters the room - Rate at which y(t) exits the room

where

Rate at which y(t) enters the room: (flow rate of gas entering) x (concentration of substance in gas entering)

Rate at which y(t) exits the room: (flow rate of gas exiting) x

(concentration of substance in gas exiting)

Fresh air enters the room at rate $2 \frac{m^3}{min}$, and 0.05% of this air is carbon dioxide, so the rate at which carbon dioxide enters the room is $2 \cdot 0.0005 = 0.001 \frac{m^3}{min}$.

The air mixture leaves the room at rate $2 \frac{m^3}{min}$, and the concentration of carbon dioxide in this air is $\frac{y}{180}$, so the rate at which carbon dioxide leaves the room is $2 \cdot \frac{y}{180}  = \frac{y}{90} \frac{m^3}{min}$.

Thus, the differential equation that governs the system is

$\frac{dy}{dt}=0.001-\frac{y}{90}$

Rearrange the above equation

$\frac{dy}{0.001-\frac{y}{90} }=dt$

Integrate both sides

$\int \frac{dy}{0.001-\frac{y}{90} }=\int dt\\\\-90\ln \left|0.001-\frac{y}{90} \right|=t+C$

Solve for y

$-90\ln \left|0.001-\frac{y}{90} \right|=t+C\\\\\ln \left|0.001-\frac{y}{90} \right|=-\frac{t}{90} +\frac{C}{90}\\\\\left|0.001-\frac{y}{90} \right|=e^{-\frac{t}{90} +\frac{C}{90}}\\\\0.001-\frac{y}{90}=\pm e^{-\frac{t}{90} +\frac{C}{90}}\\\\\frac{y}{90}=0.001-(\pm e^{-\frac{t}{90} +\frac{C}{90}})\\\\y=0.09+Ce^{-t/90}$

Now we need to use the initial condition: the room contains 0.4% carbon dioxide at time t = 0, so $y(0) = 180 \cdot 0.004 = 0.72 \:m^3$. Then

$0.72=0.09+Ce^{-0/90}\\C=0.63$

The particular solution that describes this system is

$y(t)=0.09+0.63e^{-t/90}$

where y(t) is the volume of carbon dioxide.

(A) The percentage of carbon dioxide is:

$p(t)=100\%\cdot \frac{y(t)}{180}\\\\p(t)=100\%\cdot \frac{0.09+0.63e^{-t/90}}{180}\\\\p(t)=0.55\%\left(0.63e^{-\frac{t}{90}}+0.09\right)$

(B) To find out what happens in the long run, we can take the limit as $t\rightarrow \infty$

$\lim _{t\to \infty }\left(0.55\left(0.63e^{-\frac{t}{90}}+0.09\right)\right)\\\\0.55\cdot \lim _{t\to \infty \:}\left(0.63e^{-\frac{t}{90}}+0.09\right)\\\\0.55\left(\lim _{t\to \infty \:}\left(0.63e^{-\frac{t}{90}}\right)+\lim _{t\to \infty \:}\left(0.09\right)\right)\\\\0.55\left(0+0.09\right)\\\\0.0495$

The percentage of carbon dioxide will approach 0.0495%

Answer 2

The percentage of carbon dioxide contained in the room as a function of

time is given by the rate of change of the carbon dioxide content.

Responses:

• $p(t) = \underline{\left(0.05 + 0.35\cdot e^{\dfrac{-t }{90}} \right) \%}$

• $\displaystyle \lim_{t \to \infty} p(t) = \underline{0.05\%}$

Which method can be used to find the percentage as a function of time?

Given parameters:

Volume of air in the room = 180 m³

Percentage of carbon dioxide in the air in the room = 0.4%

Percentage of carbon dioxide in air entering the room = 0.05%

Rate at which air flows into the room = 2 m³/min

Therefore;

Volume of carbon dioxide entering = 0.05/100 * 2 = 0.001 m³/min

$Concentration \ of \ CO_2 \ in \ the \ air \ leaving \ the \ room = \mathbf{ \dfrac{x}{180}}$

$Rate \ of \ CO_2 \ leaving \ the \ room = 2 \times \dfrac{x}{180} \ m^3/min = \mathbf{\dfrac{x}{90} \ m^3/min}$

Volume rate at which CO₂ leaves the room = $\frac{2 \cdot x}{180}$ = $\frac{x}{90}$

Which gives;

$\dfrac{dx}{dt} = \mathbf{0.001 - \dfrac{x}{90}}$

$\dfrac{dx}{0.001 - \dfrac{x}{90}} = dt$

$\mathbf{\displaystyle \int \dfrac{dx}{0.001 - \dfrac{x}{90}} }= \int\limits dt$

$u = {0.001 - \dfrac{x}{90}}$

$du =\mathbf{ -\dfrac{1}{90}\cdot dx}$

$\displaystyle \int \dfrac{dx}{0.001 - \dfrac{x}{90}} = -90\cdot \int \dfrac{\dfrac{1}{-90} \cdot dx}{0.001 - \dfrac{x}{90}} = {-90\cdot \int \dfrac{du}{u} =-90\cdot ln\left |0.001 - \dfrac{x}{90} \right |$

Therefore;

$\displaystyle \int \dfrac{dx}{0.001 - \dfrac{x}{90}} = -90 \cdot ln\left |{0.001 - \dfrac{x}{90}} \right | = t + c$

$\mathbf{-90 \cdot ln\left |{0.001 - \dfrac{x}{90}} \right |}= t + c$

$\left |{0.001 - \dfrac{x}{90}} \right | = \mathbf{ e^\dfrac{-t + c}{90}}$

$x = 90 \times \left(0.001 - e^{\dfrac{-t + c}{90} } \right) = 0.09 + C \cdot e^{\dfrac{-t }{90} }$

At t = 0, x = 0.004 × 180 = 0.72

Therefore;

$0.72 = 0.09 + C \cdot e^{\dfrac{0 }{90} } = 0.09 + C$

C = 0.72 - 0.09 = 0.63

Which gives;

• The volume of CO₂ as a function of time is; $f(t) = \mathbf{0.09 + 0.63 \cdot e^{\dfrac{-t }{90} }}$

Which gives;

$p(t) = \dfrac{ 0.09 + 0.63 \cdot e^{\dfrac{-t }{90} }}{180} \times 100 = \mathbf{\left(0.05 + 0.35\cdot e^{\dfrac{-t }{90}} \right) \%}$

• $p(t) = \underline{\left(0.05 + 0.35\cdot e^{\dfrac{-t }{90}} \right) \%}$

Second question;

$\displaystyle \lim_{t \to \infty} p(t) = \left(0.05 + \lim_{t \to \infty} \left(0.35\cdot e^{\dfrac{-t }{90}}\right) \right) \% = (0.05 + 0)\% =\mathbf{ 0.05\%}$

• $\displaystyle \lim_{t \to \infty} p(t) = \underline{0.05\%}$

Learn more about differential equations here:

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